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Chapter 17Free Energy
and Thermodynamics
2008, Prentice Hall
Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro
Roy KennedyMassachusetts Bay Community College
Wellesley Hills, MA
Tro, Chemistry: A Molecular Approach 2
First Law of Thermodynamics
• you can’t win!• First Law of Thermodynamics: Energy
cannot be Created or Destroyedthe total energy of the universe cannot changethough you can transfer it from one place to another
• ∆Euniverse = 0 = ∆Esystem + ∆Esurroundings
Tro, Chemistry: A Molecular Approach 3
First Law of Thermodynamics• Conservation of Energy• For an exothermic reaction, “lost” heat from the system
goes into the surroundings• two ways energy “lost” from a system,
converted to heat, qused to do work, w
• Energy conservation requires that the energy change in the system equal the heat released + work done
∆E = q + w∆E = ∆H + P∆V
• ∆E is a state functioninternal energy change independent of how done
Tro, Chemistry: A Molecular Approach 4
Energy Tax• you can’t break even!• to recharge a battery with 100 kJ of
useful energy will require more than 100 kJ
• every energy transition results in a “loss” of energy
conversion of energy to heat which is “lost” by heating up the surroundings
Tro, Chemistry: A Molecular Approach 5
Heat Taxfewer steps
generally results in a lower total
heat tax
Tro, Chemistry: A Molecular Approach 6
Thermodynamics and Spontaneity• thermodynamics predicts whether a process will
proceed under the given conditionsspontaneous process
nonspontaneous processes require energy input to go• spontaneity is determined by comparing the free
energy of the system before the reaction with the free energy of the system after reaction.
if the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable.
• spontaneity ≠ fast or slow
Tro, Chemistry: A Molecular Approach 7
Comparing Potential EnergyThe direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.
Tro, Chemistry: A Molecular Approach 8
Reversibility of Process• any spontaneous process is irreversible
it will proceed in only one direction• a reversible process will proceed back and forth
between the two end conditionsequilibriumresults in no change in free energy
• if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction
Tro, Chemistry: A Molecular Approach 9
Thermodynamics vs. Kinetics
Tro, Chemistry: A Molecular Approach 10
Diamond → Graphite
Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in
your lifetime (or through many of your generations).
Tro, Chemistry: A Molecular Approach 11
Factors Affecting Whether a Reaction Is Spontaneous
• The two factors that determine the thermodynamic favorability are the enthalpy and the entropy.
• The enthalpy is a comparison of the bond energy of the reactants to the products.
bond energy = amount needed to break a bond. ∆H
• The entropy factors relates to the randomness/orderliness of a system
∆S• The enthalpy factor is generally more important
than the entropy factor
Tro, Chemistry: A Molecular Approach 12
Enthalpy• related to the internal energy• ∆H generally kJ/mol• stronger bonds = more stable molecules• if products more stable than reactants, energy released
exothermic∆H = negative
• if reactants more stable than products, energy absorbedendothermic∆H = positive
• The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions.
• Hess’ Law ∆H°rxn = Σ(∆H°prod) - Σ(∆H°react)
Substance ∆H° kJ/mol
Substance ∆H° kJ/mol
Al(s) 0 Al2O3 -1669.8 Br2(l) 0 Br2(g) +30.71
C(diamond) +1.88 C(graphite) 0 CO(g) -110.5 CO2(g) -393.5 Ca(s) 0 CaO(s) -635.5 Cu(s) 0 CuO(s) -156.1 Fe(s) 0 Fe2O3(s) -822.16 H2(g) 0 H2O2(l) -187.8
H2O(g) -241.82 H2O(l) -285.83 HF(g) -268.61 HCl(g) -92.30
HBr(g) -36.23 HI(g) +25.94 I2(s) 0 I2(g) +62.25 N2(g) 0 NH3(g) -46.19 NO(g) +90.37 NO2(g) +33.84 Na(s) 0 O2(g) 0 S(s) 0 SO2(g) -296.9
Tro, Chemistry: A Molecular Approach 14
Entropy• entropy is a thermodynamic function that increases as the
number of energetically equivalent ways of arranging the components increases, S• S generally J/mol
• S = k ln Wk = Boltzmann Constant = 1.38 x 10-23 J/KW is the number of energetically equivalent ways, unitless
• Random systems require less energy than ordered systems
Tro, Chemistry: A Molecular Approach 15
W
Energetically Equivalent States for the Expansion of a Gas
Tro, Chemistry: A Molecular Approach 16
Macrostates → Microstates
This macrostate can be achieved throughseveral different arrangements of the particles
These microstates all have the same
macrostate
So there are 6 different particle arrangements that result in the same
macrostate
Tro, Chemistry: A Molecular Approach 17
Macrostates and ProbabilityThere is only one possible
arrangement that gives State A and one that gives State C
There are 6 possible arrangements that give State B
Therefore State B has higher entropy than either
State A or State B
The macrostate with the highest entropy also has the greatest dispersal of energy
Tro, Chemistry: A Molecular Approach 18
Changes in Entropy, ∆S• entropy change is favorable when the result is a more
random system. ∆S is positive
• Some changes that increase the entropy are:reactions whose products are in a more disordered state.
(solid > liquid > gas)reactions which have larger numbers of product molecules than reactant molecules.increase in temperaturesolids dissociating into ions upon dissolving
Tro, Chemistry: A Molecular Approach 19
Increases in Entropy
Tro, Chemistry: A Molecular Approach 20
The 2nd Law of Thermodynamics• the total entropy change of the universe must be
positive for a process to be spontaneous for reversible process ∆Suniv = 0, for irreversible (spontaneous) process ∆Suniv > 0
• ∆Suniverse = ∆Ssystem + ∆Ssurroundings• if the entropy of the system decreases, then the
entropy of the surroundings must increase by a larger amount
when ∆Ssystem is negative, ∆Ssurroundings is positive• the increase in ∆Ssurroundings often comes from the
heat released in an exothermic reaction
Tro, Chemistry: A Molecular Approach 21
Entropy Change in State Change• when materials change state, the number of
macrostates it can have changes as wellfor entropy: solid < liquid < gasbecause the degrees of freedom of motion increases solid → liquid → gas
Tro, Chemistry: A Molecular Approach 22
Entropy Change and State Change
Tro, Chemistry: A Molecular Approach 23
Heat Flow, Entropy, and the 2nd Law
Heat must flow from water to ice in order for the entropy of the universe to increase
Tro, Chemistry: A Molecular Approach 24
Temperature Dependence of ∆Ssurroundings
• when a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings
• when a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings
• the amount the entropy of the surroundings changes depends on the temperature it is at originally
the higher the original temperature, the less effect addition orremoval of heat has
TH
S systemgssurroundin
∆−=∆
Tro, Chemistry: A Molecular Approach 25
Gibbs Free Energy, ∆G• maximum amount of energy from the system
available to do work on the surroundingsG = H – T·S
∆Gsys = ∆Hsys – T∆Ssys∆Gsys = – T∆Suniverse
∆Greaction = Σ n∆Gprod – Σ n∆Greact• when ∆G < 0, there is a decrease in free
energy of the system that is released into the surroundings; therefore a process will be spontaneous when ∆G is negative
( )
KJ3
KkJ
surr
syssurr
106.86 86.6SK 298
kJ 2044TH
S
×==∆
−−=
∆−=∆
Ex. 17.2a – The reaction C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) has ∆Hrxn = -2044 kJ at 25°C.
Calculate the entropy change of the surroundings.
combustion is largely exothermic, so the entropy of the surrounding should increase significantly
∆Hsystem = -2044 kJ, T = 298 K
∆Ssurroundings, J/K
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
∆ST, ∆H
TH
S syssurr
∆−=∆
Tro, Chemistry: A Molecular Approach 27
Free Energy Change and Spontaneity
Tro, Chemistry: A Molecular Approach 28
Gibbs Free Energy, ∆G• process will be spontaneous when ∆G is negative• ∆G will be negative when
∆H is negative and ∆S is positiveexothermic and more random
∆H is negative and large and ∆S is negative but small∆H is positive but small and ∆S is positive and
largeor high temperature
• ∆G will be positive when ∆H is + and ∆S is −never spontaneous at any temperature
• when ∆G = 0 the reaction is at equilibrium
Tro, Chemistry: A Molecular Approach 29
∆G, ∆H, and ∆S
Ex. 17.3a – The reaction CCl4(g) → C(s, graphite) + 2 Cl2(g) has ∆H = +95.7 kJ and ∆S = +142.2 J/K at 25°C.
Calculate ∆G and determine if it is spontaneous.
Since ∆G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.
∆H = +95.7 kJ, ∆S = 142.2 J/K, T = 298 K
∆G, kJ
Answer:
Solution:
Concept Plan:
Relationships:
Given:
Find:
∆GT, ∆H, ∆S
STHG ∆−∆=∆
( ) ( )( )J1033.5
2.142K 298J 1095.7
STHG
4KJ3
×+=
+−×+=
∆−∆=∆
Ex. 17.3a – The reaction CCl4(g) → C(s, graphite) + 2 Cl2(g) has ∆H = +95.7 kJ and ∆S = +142.2 J/K.
Calculate the minimum temperature it will be spontaneous.
The temperature must be higher than 673K for the reaction to be spontaneous
∆H = +95.7 kJ, ∆S = 142.2 J/K, ∆G < 0
Τ, Κ
Answer:
Solution:
Concept Plan:
Relationships:
Given:
Find:
T∆G, ∆H, ∆S
STHG ∆−∆=∆
( )( ) ( )( )( ) ( )( )K
J3KJ3
2.142TJ 1095.7
02.142TJ 1095.7
0 STHG
+<×+
<+−×+
<∆−∆=∆ ( )( )
TK 673
T 2.142
J 1095.7
KJ
3
<
<+
×+
Tro, Chemistry: A Molecular Approach 32
The 3rd Law of ThermodynamicsAbsolute Entropy
• the absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles
• the 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol·K
therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropytherefore, the absolute entropy of substances is always +
Tro, Chemistry: A Molecular Approach 33
Standard Entropies
• S°• extensive• entropies for 1 mole at 298 K for a particular
state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution
Substance S° J/mol-K
Substance S° J/mol-K
Al(s) 28.3 Al2O3(s) 51.00 Br2(l) 152.3 Br2(g) 245.3
C(diamond) 2.43 C(graphite) 5.69 CO(g) 197.9 CO2(g) 213.6 Ca(s) 41.4 CaO(s) 39.75 Cu(s) 33.30 CuO(s) 42.59 Fe(s) 27.15 Fe2O3(s) 89.96 H2(g) 130.58 H2O2(l) 109.6
H2O(g) 188.83 H2O(l) 69.91 HF(g) 173.51 HCl(g) 186.69
HBr(g) 198.49 HI(g) 206.3 I2(s) 116.73 I2(g) 260.57 N2(g) 191.50 NH3(g) 192.5 NO(g) 210.62 NO2(g) 240.45 Na(s) 51.45 O2(g) 205.0 S(s) 31.88 SO2(g) 248.5
Tro, Chemistry: A Molecular Approach 35
Relative Standard EntropiesStates
• the gas state has a larger entropy than the liquid state at a particular temperature
• the liquid state has a larger entropy than the solid state at a particular temperature
188.8H2O (l)70.0H2O (g)
S°, (J/mol·K)
Substance
Tro, Chemistry: A Molecular Approach 36
Relative Standard EntropiesMolar Mass
• the larger the molar mass, the larger the entropy
• available energy states more closely spaced, allowing more dispersal of energy through the states
Tro, Chemistry: A Molecular Approach 37
Relative Standard EntropiesAllotropes
• the less constrained the structure of an allotrope is, the larger its entropy
Tro, Chemistry: A Molecular Approach 38
Relative Standard EntropiesMolecular Complexity
• larger, more complex molecules generally have larger entropy
• more available energy states, allowing more dispersal of energy through the states
30.00639.948
MolarMass
210.8NO (g)154.8Ar (g)
S°, (J/mol·K)
Substance
Tro, Chemistry: A Molecular Approach 39
Relative Standard EntropiesDissolution
• dissolved solids generally have larger entropy
• distributing particles throughout the mixture
265.7KClO3(aq)143.1KClO3(s)
S°, (J/mol·K)
Substance
Ex. 17.4 –Calculate ∆S° for the reaction4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l)
∆S is +, as you would expect for a reaction with more gas product molecules than reactant molecules
standard entropies from Appendix IIB∆S, J/K
Check:
Solution:
Concept Plan:
Relationships:
Given:Find:
∆SS°NH3, S°O2, S°NO, S°H2O,
( ) ( )reactantsproducts SSS ooorp nn Σ−Σ=∆
( ) ( )
KJ
KJ
KJ
KJ
KJ
)(O)(NH)O(H)NO(
reactantsproducts
8.178
)]2.205(5)8.192(4[)]8.188(6) 8.210(4[
)]S(5)S(4[)]S(6)S(4[
SSS
232
=
+−+=
+−+=
Σ−Σ=∆
gggg
rp nnoooo
ooo
188.8H2O(g)210.8NO(g)205.2O2(g)192.8NH3(g)
S°, J/mol⋅KSubstance
Tro, Chemistry: A Molecular Approach 41
Calculating ∆G°
• at 25°C:∆Go
reaction = ΣnGof(products) - ΣnGo
f(reactants)
• at temperatures other than 25°C:assuming the change in ∆Ho
reaction and ∆Soreaction is
negligible∆G°reaction = ∆H°reaction – T∆S°reaction
42
Substance ∆G°f kJ/mol
Substance ∆G°f kJ/mol
Al(s) 0 Al2O3 -1576.5 Br2(l) 0 Br2(g) +3.14
C(diamond) +2.84 C(graphite) 0 CO(g) -137.2 CO2(g) -394.4 Ca(s) 0 CaO(s) -604.17 Cu(s) 0 CuO(s) -128.3 Fe(s) 0 Fe2O3(s) -740.98 H2(g) 0 H2O2(l) -120.4
H2O(g) -228.57 H2O(l) -237.13 HF(g) -270.70 HCl(g) -95.27
HBr(g) -53.22 HI(g) +1.30 I2(s) 0 I2(g) +19.37 N2(g) 0 NH3(g) -16.66 NO(g) +86.71 NO2(g) +51.84 Na(s) 0 O2(g) 0 S(s) 0 SO2(g) -300.4
Ex. 17.7 –Calculate ∆G° at 25°C for the reaction
CH4(g) + 8 O2(g) → CO2(g) + 2 H2O(g) + 4 O3(g)
standard free energies of formation from Appendix IIB∆G°, kJ
Solution:
Concept Plan:
Relationships:
Given:Find:
∆G°∆G°f of prod & react
( ) ( )reactantsproducts GGG ooofrfp nn ∆Σ−∆Σ=∆
( ) ( )
kJ 3.148)]kJ 0.0(8)kJ 5.50[(]kJ) 2.163()kJ 6.228(2)kJ 4.394[(
)]G(8)G[()]G()G(2)G[(
GGG
24322 OCHOOHCO
reactantsproducts
−=+−−++−+−=
∆+∆−∆+∆+∆=
∆Σ−∆Σ=∆ooooo
ooo
fffff
frfp nn
163.2O3(g)-228.6H2O(g)-394.4CO2(g)
0.0O2(g)-50.5CH4(g)
∆G°f, kJ/molSubstance
Ex. 17.6 – The reaction SO2(g) + ½ O2(g) → SO3(g) has ∆H° = -98.9 kJ and ∆S° = -94.0 J/K at 25°C.
Calculate ∆G° at 125°C and determine if it is spontaneous.
Since ∆G is -, the reaction is spontaneous at this temperature, though less so than at 25°C
∆H° = -98.9 kJ, ∆S° = -94.0 J/K, T = 398 K
∆G°, kJ
Answer:
Solution:
Concept Plan:
Relationships:
Given:
Find:
∆G°T, ∆H°, ∆S°ooo STHG ∆−∆=∆
( ) ( )( )kJ5.61J105.61
0.94K 398J 1098.9
STHG
3KJ3
−=×−=
−−×−=
∆−∆=∆ ooo
Tro, Chemistry: A Molecular Approach 45
∆G Relationships• if a reaction can be expressed as a series of reactions,
the sum of the ∆G values of the individual reaction is the ∆G of the total reaction
∆G is a state function
• if a reaction is reversed, the sign of its ∆G value reverses
• if the amounts of materials is multiplied by a factor, the value of the ∆G is multiplied by the same factor
the value of ∆G of a reaction is extensive
Tro, Chemistry: A Molecular Approach 46
Free Energy and Reversible Reactions• the change in free energy is a theoretical limit as
to the amount of work that can be done• if the reaction achieves its theoretical limit, it is
a reversible reaction
Tro, Chemistry: A Molecular Approach 47
Real Reactions• in a real reaction, some of the free energy is
“lost” as heatif not most
• therefore, real reactions are irreversible
Tro, Chemistry: A Molecular Approach 48
∆G under Nonstandard Conditions⋅ ∆G = ∆G° only when the reactants and products
are in their standard states⋅ there normal state at that temperature⋅ partial pressure of gas = 1 atm⋅ concentration = 1 M
⋅ under nonstandard conditions, ∆G = ∆G° + RTlnQ⋅ Q is the reaction quotient
⋅ at equilibrium ∆G = 0⋅ ∆G° = ─RTlnK
Tro, Chemistry: A Molecular Approach 49
50
Example - ∆G• Calculate ∆G at 427°C for the reaction below if the
PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atmN2(g) + 3 H2(g) → 2 NH3(g)
Q = PNH3
2
PN21 x PH2
3
(2.0 atm)2
(33.0 atm)1 (99.0)3= = 1.2 x 10-7
∆G = ∆G° + RTlnQ∆G = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
∆G = -46300 J = -46 kJ
∆H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
∆S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K∆G° = -92380 J - (700 K)(-198.2 J/K)
∆G° = +46400 J
Tro, Chemistry: A Molecular Approach 51
52
Example - K• Estimate the equilibrium constant and position of
equilibrium for the following reaction at 427°CN2(g) + 3 H2(g) ⇔ 2 NH3(g)
∆G° = -RT lnK+46400 J = -(8.314 J/K)(700 K) lnK
lnK = -7.97K = e-7.97 = 3.45 x 10-4
since K is << 1, the position of equilibrium favors reactants
∆H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J∆S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
∆G° = -92380 J - (700 K)(-198.2 J/K)∆G° = +46400 J
Tro, Chemistry: A Molecular Approach 53
Temperature Dependence of K• for an exothermic reaction, increasing the
temperature decreases the value of the equilibrium constant
• for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant
RS
T1
RHln rxnrxn
oo ∆+
∆
−=K
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